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Q.

A geo-stationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R  being the radius of the earth. The time period of another satellite (in hours) at a height of 2R  from the surface of the earth is [CBSE AIPMT 2012]

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a

5

b

10

c

62

d

6/2

answer is C.

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Detailed Solution

From Kepler's third law,   T2∝r3Hence,  T12∝r13 and T22∝r23So,      T22T12=r23r13=(3R)3(6R)3 or T22T12=18⇒  T22=18T12⇒  T2=2422=62 h
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